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Alex787 [66]
3 years ago
7

Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calcula

tions and the glassware used to perform the dilution.)
Chemistry
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

<em> = 0.2 mL.</em>

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

C_{1} V_{1} = C_{2} V_{2}

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= \frac{10}{0.5} × 0.010

<em> = 0.2 mL.</em>

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

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Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You
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Answer:

0.11%

Explanation:

Without mincing words, let us dive straight into the solution to the question/problem. The first step to solve this question is to write out the chemical reaction, that is the reaction showing the dissociation of acetic acid.

CH3COOH <=======================================> CH3COO⁻ + H⁺

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The next thing to do is to calculate for the percentage of dissociation, this can be done as given below:

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7 0
2 years ago
I don’t get it plz help
d1i1m1o1n [39]
D=m/v. (Density equals mass over volume).
So in your case density = 4.2/6
Which equals 0.7
Your units will be grams over milliliters

So your answer is 0.7 g/ml
3 0
3 years ago
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